Question

The pH of $0.3M$ acetic acid $\left(C{H}_{3}COOH\right)$ with $K_a=1.8\times {10}^{-5}$ is :
Take $lo{g}_{10}\left(2.32\right)=0.36$

• A. 6.56
• B. 2.64
• C. 4.86
• D. 7.63

Answer 1

The correct option is B 2.64

$K_a=1.8\times {10}^{-5}$
the dissociation of acetic acid is given by
$C{H}_{3}COOH\rightleftharpoons H^{+}+C{H}_{3}CO{O}^{-}Initial:0.300Atequilibrium:0.3-xxxHerexisthedissociationofC{H}_{3}COOHatequilibrium$

$\Rightarrow K_a=1.8\times {10}^{-5}=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}=\frac{x\times x}{0.3-x}SinceC{H}_{3}COOHisaweakacidx<<0.3$
$\therefore 0.3-x\approx 0.3K_a=1.8\times {10}^{-5}=\frac{x^2}{0.3}\Rightarrow x=2.32\times {10}^{-3}x=\left[H^{+}\right]=2.32\times {10}^{-3}pH=-lo{g}_{10}\left[H^{+}\right]pH=-lo{g}_{10}(2.3\times {10}^{-3})pH=2.638$